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Question Number 156677 by saly last updated on 14/Oct/21

Commented by cortano last updated on 14/Oct/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt[\mathrm{n}]{\mathrm{a}}+\sqrt[\mathrm{n}]{\mathrm{b}}}{2}\right)}^{\mathrm{n}}?$$

Commented by saly last updated on 14/Oct/21

$$whath?$$

Commented by john_santu last updated on 14/Oct/21

$$ifA=\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)}^n={\left(\frac{\sqrt[n]{a}+\sqrt[b]{n}}{2}\right)}^n\times \underset{x\to \mathrm{\infty }}{\lim }\left(1\right)$$$$={\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)}^n$$

Answered by puissant last updated on 14/Oct/21

$$A=\underset{n\to +\mathrm{\infty }}{\lim }{\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)}^n;x=\frac{1}{n}\to n=\frac{1}{x}$$$$A=\underset{x\to 0}{\lim }{\left(\frac{a^x+b^x}{2}\right)}^{\frac{1}{x}}=\underset{x\to 0}{\lim }{\left(\frac{1+xlna+1+xlnb}{2}\right)}^{\frac{1}{x}}$$$$=\underset{x\to 0}{\lim }{(1+\frac{x}{2}ln\left(ab\right))}^{\frac{1}{x}}=\underset{x\to 0}{\lim }{e}^{\frac{ln(1+\frac{x}{2}lnab)}{x}}$$$$=\underset{x\to 0}{\lim }{e}^{\frac{\frac{x}{2}lnab}{x}}=e^{\frac{1}{2}lnab}=\sqrt{ab}...$$$$\therefore \because A=\underset{n\to +\mathrm{\infty }}{\lim }{\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)}^n=\sqrt{ab}....$$$$$$$$............\mathcal{L}epuissant............$$

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